Question 789080
{{{ sqrt(3)(cot(theta))^2 - 4cot(theta) + sqrt(3)= 0 }}}
Let's make a change of varaible so I do not have to write {{{cot(theta)}}} so many times.
{{{cot(theta)=y}}}
Now I can write the equation as
{{{sqrt(3)y^2-4y+sqrt(3)=0}}}
We have to solve that equation
It may not be obvious, but we can solve that equation by factoring.
The left diside of the equal sign can be factored to get
{{{(sqrt(3)y-1)(y-sqrt(3))=0}}}
That leads to 2 solutions
{{{sqrt(3)y-1=0}}}<-->{{{y=1/sqrt(3)=sqrt(3)/3}}}
{{{y-sqrt(3)=0}}}<-->{{{y=sqrt(3)}}}
Since {{{cot(theta)=1/tan(theta)}}}, the solutions look very symmetrical:
{{{cot(theta)=1/sqrt(3)}}}-->{{{tan(theta)=1/cot(theta)=sqrt(3)}}}
and
{{{cot(theta)=sqrt(3)}}}-->{{{tan(theta)=1/cot(theta)=1/sqrt(3)}}}
One function is {{{sqrt(3)}}} and the other is {{{1/sqrt(3)=sqrt(3)/3}}}.
For either set of solutions,
{{{(cot(theta))^2+(tan(theta))^2=(sqrt(3))^2+(sqrt(3)/3)^2=3+3/9=3+1/3=highlight(10/3)}}}
 
PS _ No, I would not have tried to use the quadratic formula to solve
{{{sqrt(3)y^2-4y+sqrt(3)=0}}}, but I did not think of the factorization right away.
The way I really figured out the solutions was dividing both sides by {{{sqrt(3)}}} (same as multiplying times {{{sqrt(3)/3}}} if you like it better that way, and then "completing the square:
{{{sqrt(3)y^2-4y+sqrt(3)=0}}}
{{{y^2-(4/sqrt(3))y+1=0}}}
{{{y^2-2(2/sqrt(3))y=-1}}}
{{{y^2-2(2/sqrt(3))y+(2/sqrt(3))^2=-1+(2/sqrt(3))^2}}}
{{{(y-2/sqrt(3))^2=-1+4/3}}}
{{{(y-2/sqrt(3))^2=1/3}}}
Then, taking square roots:
{{{system(y-2/sqrt(3)=1/sqrt(3),"or",y-2/sqrt(3)=-1/sqrt(3))}}} --> {{{system(y=2/sqrt(3)+1/sqrt(3),"or",y=2/sqrt(3)-1/sqrt(3))}}} --> {{{system(y=3/sqrt(3),"or",y=1/sqrt(3))}}}
 
EXTRA:
The problem did not ask about the angles, but the angles in the first quadrant that have those tangent and cotangent values are {{{pi/6}}} and {{{pi/3}}} (or {{{30^o}}} and {{{60^o}}} if you prefer degrees).